Assume that you have an aqueous mixture of BaCl2 and CuCl2. How could you use a precipitation reaction to sepa-rate the two metal ions?

Verified step by step guidance

1. Identify the ions present in the solution: Ba^{2+}, Cu^{2+}, and Cl^{-}.
2. Choose a reagent that will selectively precipitate one of the metal ions. In this case, use a sulfate source like Na2SO4, which will precipitate Ba^{2+} as BaSO4.
3. Add the Na2SO4 solution to the mixture. BaSO4 will form as a solid precipitate, while Cu^{2+} will remain in solution.
4. Filter the mixture to separate the solid BaSO4 from the aqueous solution containing Cu^{2+}.
5. Wash the precipitate to remove any remaining Cu^{2+} ions and dry the BaSO4 precipitate.

Key Concepts

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form an insoluble solid, known as a precipitate. This process is driven by the formation of a compound that has low solubility in the solvent, leading to its separation from the solution. Understanding the solubility rules helps predict which combinations of ions will form a precipitate.

Solubility Rules

Solubility rules are guidelines that help predict whether a compound will dissolve in water. For instance, most chlorides are soluble, except for those of silver, lead, and mercury. Barium compounds, on the other hand, are generally insoluble, which is crucial for separating Ba²⁺ from Cu²⁺ in the given mixture through selective precipitation.

Selective Precipitation

Selective precipitation is a technique used to separate ions in a solution by adding a reagent that forms a precipitate with one ion while leaving others in solution. In this case, adding a sulfate source can precipitate barium sulfate (BaSO4), allowing for the separation of Ba²⁺ from Cu²⁺, which remains soluble. This method is effective for isolating specific metal ions based on their differing solubility.